There is a group of five children, where two of the children are twins. In how many ways can I distribute 8 identical pieces of candy to the children, if the twins must get an equal amount of candy?

Lilliam0216 Jun 8, 2024

#1**0 **

There are 8 candies to distribute, and the twins must get the same amount. So, let's say they each get x candies (where x can be any number from 0 to 4).

This leaves 8 - 2x candies for the remaining 3 children. We don't care about the order we give candy to these 3 children, so we can focus on how many candies each gets. There are 3 + (8-2x) - 1 = 9 - 2x candies to distribute among these 3 children, where everyone gets at least 1 (because 9-2x >= 3).

We can use stars and bars to solve this kind of problem. Imagine we have 9 - 2x stars and 2 indistinguishable bars. Placing the bars divides the stars into 3 groups representing the candy allocation for the 3 children (excluding the twins).

For example, if the twins each get 2 candies (x = 2), we have 5 stars and 2 bars. Here, ||** represents 2 candies for the first child, 0 candies for the second child, and 3 candies for the third child.

The number of arrangements of stars and bars is the same as the number of ways to distribute the candies. This can be calculated using the formula for combinations: nCr = n! / (r! * (n-r)!), where n is the total number of objects (stars + bars) and r is the number of bars.

In our case, n = (9 - 2x) + 2 = 11 - 2x and r = 2. So, the number of arrangements for a given value of x is:

(11 - 2x)! / (2! * (11 - 2x - 2)!) = (11 - 2x)! / (2! * (9 - 2x)!)

Now we need to consider all the possible values of x (from 0 to 4) and add up the number of arrangements for each case. This gives us the total number of ways to distribute the candy.

However, there's a shortcut! We realize that the arrangements for a given value of x will be the same as the arrangements for 4 - x (because the stars and bars are indistinguishable). For example, the arrangements for x = 2 (as shown above) are the same as the arrangements for x = 4 (where ||** becomes **||).

Therefore, we only need to consider values of x from 0 to 2 and then multiply the final result by 2 (to account for the double counting).

So, summing the arrangements for x = 0, 1, and 2:

x = 0: (11 - 0)! / (2! * (9 - 0)!) = 11!/ (2! * 9!) = 55

x = 1: (11 - 2)! / (2! * (9 - 2)!) = 9!/ (2! * 7!) = 36

x = 2: (11 - 4)! / (2! * (9 - 4)!) = 7!/ (2! * 5!) = 21

Total arrangements (considering double counting): 2 * (55 + 36 + 21) = 2 * 112 = 224

However, there's one last detail. We've counted the arrangements where the twins get 0 candy. But since they must get at least one candy each, we need to subtract the arrangements where x = 0 (which is 55).

Therefore, the final number of ways to distribute the candy is 224 - 55 = 169 ways.

bader Jun 9, 2024